A question on summable families in topological groups I want to prove the following result involving summable families in abelian $T_2$ topological groups.
Let $u:G\to G'$ be a continuous group homomorphism between two abelian $T_2$ topological group, which is a local homeomorphism at $0$ i.e. $\exists V,V'$ neighborhoods of $0$ such that $u|_V:G|_V\to G'|_{V'}$ is a homeomorphism.
Now, take a family $(x_i)_{i \in I}$ inside $G$ such that given any neighborhood $W$ of $0$, all but finitely many $x_i's$ lie inside $W$.
Then $(x_i)_{i\in I}$ is a summable family inside $G$ if and only if $(u(x_i))_{i\in I}$ is summable inside $G'$.
I would just like to mention the definition of summability that we are using here.
Definition of summability in an abelian $T_2$ topological group G: A family $(x_i)_{i\in I}$ is said to be summable to $s\in G$  if for any given neighborhood $V$ of $s$, $\exists J_0$, a finite subset of $I$ such that $\forall J_0\subset J\subset I$, $J$ finite,  $\sum_{i\in J} x_i\in V$.
One more definition: A family $(x_i)_{i\in I}$ is said to satisfy Cauchy criteria if for any given neighborhood $V$ of $0$,$\exists J_0$, a finite subset of $I$ such that $\forall K\subset I$;$|K|<\infty$;$J_0\cap K=\phi$,   $\sum_{i\in K} x_i\in V$.
Now, I have been able to show that if the family $(x_i)_{i\in I}$ is summable inside $G$,then the family  $(u(x_i))_{i\in I}$ is also summable inside $G'$. But I am stuck with the converse part. Any help would be appreciated. Thanks in advance.
My approach: Since the definition of local homeomorphism involves a neighborhood of $0$, I was trying to use the Cauchy criteria for a family. But Cauchy criteria may not imply summability if the given space is not complete.
Also, I am new to the theory of topological groups. I do not know too many results. So, if the answer is self-contained , that would be really helpful to  me.
 A: Suppose first that the family $(x_i)_{i \in I}$ is summable in $G,$
with sum $s.$ Let $V$ be any neighbourhood of $0$ in $G',$ and
define $U = u^{-1}(V).$ Then $U$ is a neighbourhood of $0$ in $G.$
By the summability of $(x_i)_{i \in I},$ there exists a finite set
$J \subseteq I$ such that whenever $K$ is a finite set satisfying
$J \subseteq K \subseteq I,$
$$
\sum_{i \in K}x_i \in s + U.
$$
For all such $K,$
$$
\sum_{i \in K}u(x_i) = u\bigg(\sum_{i \in K}x_i\bigg)
\in  u(s + U) = u(s) + V.
$$
Therefore the family $(u(x_i))_{i \in I}$ is summable in $G',$ with
sum $u(s).$
So far, we haven't used all of the hypotheses of the question; but,
apart from the $\text{T}_2$ (Hausdorff) hypothesis, they are all
needed for the converse.
Suppose now that the family $(u(x_i))_{i \in I}$ is
summable in $G',$ with sum $s'.$ Let $A$ be a neighbourhood of $0$
in $G$ and $B$ a neighbourhood of $0$ in $G'$ such that $u$
restricts to a homeomorphism $u|_{A, B} \colon A \to B.$
By the hypothesis about the family $(x_i)_{i \in I}$
[I don't know if there is a term for this condition, incidentally,
but I suppose one could say that the family is "mostly small"],
there exists a finite
set $M \subseteq I$ such that $x_i \in A$ for all
$i \in I \setminus M.$ By the summability of $(u(x_i))_{i \in I},$
there exists a finite set $N \subseteq I,$ which without loss of
generality we may suppose contains $M,$ such that whenever $K$ is a
finite set satisfying $N \subseteq K \subseteq I,$
$$
\sum_{i \in K}u(x_i) \in s' + B.
$$
Define
$$
s = \sum_{i \in N}x_i +
(u|_{A, B})^{-1}\bigg(s' - \sum_{i \in N}u(x_i)\bigg).
$$
We will show that $(x_i)_{i \in I}$ is summable in $G,$ with
sum $s.$
If $K$ is any finite set satisfying $N \subseteq K \subseteq I,$
then $K \setminus N \subseteq I \setminus M,$ therefore
$x_i = (u|_{A, B})^{-1}(u(x_i))$ for all $i \in K \setminus N,$
therefore
$$
\sum_{i \in K}x_i - s =
\sum_{i \in K \setminus N}x_i + \sum_{i \in N}x_i - s =
(u|_{A, B})^{-1}\bigg(\sum_{i \in K}u(x_i) - s'\bigg).
$$
Let $U$ be any neighbourhood of $0$ in $G,$ and define
$V = u(U \cap A).$ Then $V$ is a neighbourhood of $0$ in $G',$ $V$
is contained in $B,$ and $u$ maps $s + (U \cap A)$ to $s' + V.$ By
the summability of $(u(x_i))_{i \in I},$ there exists a finite set
$J \subseteq I,$ which  without loss of generality we may suppose
contains $N,$ such that whenever $K$ is a finite set satisfying
$J \subseteq K \subseteq I,$
$$
\sum_{i \in K}u(x_i) - s' \in V.
$$
Therefore
$$
\sum_{i \in K}x_i - s \in (u|_{A, B})^{-1}(V) = U \cap A
\subseteq U,
$$
and this completes the proof.
